We Keep Coding

why is the error thinking Its not string?

path of file is:
"C:\Users\deana\OneDrive\Marlon's files\Programming\Python\PITT\PITT_LIbrary\Lists\test.txt"
lines of code are:
import os
os.chdir("C:/Users/deana/OneDrive/Marlon's files/Programming/Python/PITT/PITT_LIbrary/Lists")
exec(open('test.txt'))
the error is this:
Traceback (most recent call last):
File "<pyshell#14>", line 1, in <module>
exec(open('test.txt'))
TypeError: exec() arg 1 must be a string, bytes or code object
also if I try on one line as such:
exec(open(r"C:/Users/deana/OneDrive/Marlon's files/Programming/Python/PITT/PITT_LIbrary/Lists/test.txt"))
i'ts the same error. (with and without r)
super frustrationg as it reads like i'm not inputting string... but it is string!?!
also I've done this litteraly the same way before, restarted IDLE shell, no difference.
ugh! I always get stupid errors with file paths.

I should have been using os.startfile() to open this.
It was confusing by using .open(). as I was attempting to open in default app.
before, i've used exec.open() to open .py files and guess I got them confused.
exec is just used to open other scripts... need stronger coffee next time.

Try this:
import os
os.chdir("C:/Users/deana/OneDrive/Marlon's files/Programming/Python/PITT/PITT_LIbrary/Lists")
exec(open('test.txt', 'rb'))
You can convert the txt file to bytes by opening it with rb (read bytes).

How to fix this error while executing a program

C:\Users\skandregula\AppData\Local\Programs\Python\Python37\python.exe
C:/Users/skandregula/Desktop/TestFiles2/testing.py
Traceback (most recent call last):
File "C:/Users/skandregula/Desktop/TestFiles2/testing.py", line 30, in <module>
with open(in_dir + f, 'r') as tmp_file:
FileNotFoundError: [Errno 2] No such file or directory:
'C:/Users/skandregula/Desktop/TestFiles2/history.log.3.3C'
Process finished with exit code 1
that is the error i am running right now...How to fix this?

Without seeing what your code looks like, I'm assuming somewhere in your code you're referencing a file or folder named 'history.log.3.3C' as the error suggests, and that it cannot find it in your directory given. My best advice is make sure the directory is correct?
Hard to tell without seeing what you're trying to do. Maybe posting a snippet of your code would help.

Is this the right way to use open() in python? its showing an error

Using the code below:
txt = open("/data/cards/deckofcards.txt").read().splitlines()
I get the following error:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IOError: [Errno 2] No such file or directory: '/data/cards/deckofcards.txt'

txt=open("/home/vamsi/data/cards/deckofcards.txt").read().splitlines()
print(v)
I missed the (/home/vamsi) path.I misunderstood that pwd is by default and no need to give it in open().I figured it out and it is working fine.in open() we should also give pwd also to open the (.txt,.tsv etc..)file as we needed.
thank you,
screen shot of terminal

Give the absolute path of the file.
It would be something like C:/Desktop/data/card.txt.
The directory you mentioned does not exist and hence it is giving you that error.

How to load DICOM files in Python?

I am trying to figure out the basics of importing DICOM files in Python using pydicom. While trying really simple code, I get following errors:
For code:
import dicom
filePath="C:\Python34\Lib\site-packages\dicom\testfiles"
ds=dicom.read_file(filePath[0])
I get error:
C:\Python34\python.exe C:/Users/041213/PycharmProjects/D/Deki.py
Traceback (most recent call last):
File "C:/Users/041213/PycharmProjects/D/Deki.py", line 4, in
ds=dicom.read_file(filePath[0])
File "C:\Python34\lib\site-packages\dicom\filereader.py", line 589, in read_file
fp = open(fp, 'rb')
FileNotFoundError: [Errno 2] No such file or directory: 'C'
I am using Python 3.4, pydicom 0.9.9 and JetBrains PyCharm Community Edition 2016.3.2
If anyone can help me with this, or even just help me how to load a DICOM file in general, I would appreciate it a lot.

you're passing the first char of the string (C) instead of the full string. Just do:
ds=dicom.read_file(filePath)
next error you'll stumble into: use raw prefix or \t gets interpreted as a tabulation character:
filePath=r"C:\Python34\Lib\site-packages\dicom\testfiles"
^

Use and \ escape character to avoid issues with tab and other special characters. Also remember when you do filePath[0] on a string it returns the first character
filePath="C:\\Python34\\Lib\\site-packages\\dicom\\testfiles"
ds=dicom.read_file(filePath)

IOError with Python and OpenCV

I want to do some OpenCV Basic Operations using Python. My Problem is that the Pythoninterpreter says that the file I want to open with cv.LoadImage() dont exists.
But as you can see in my code and the and the Interpreter Output this file exists and the Program should be able to read it.
Likly the answer is simple (I am new at Python Programming!).
Thanks for answers!
Here my Code:
import cv, sys, os
print sys.argv[1]
print os.getcwd()
print os.access(sys.argv[1], os.F_OK)
img = cv.LoadImage(sys.argv[1], 1)
cv.NamedWindow("orginal", CV_WINDOW_AUTOSIZE)
cv.ShowImage("orginal", img)
cv.waitKey(0)
here is the Pythoninterpreter Output:
dennis#Powertux:~/opencv/showPicture$ python2.5 showPicture.py google-de02.jpg google-de02.jpg
/home/steffke/opencv/showPicture
True
Traceback (most recent call last):
File "showPicture.py", line 7, in <module>
img = cv.LoadImage(sys.argv[1], 1)
IOError: [Errno 2] No such file or directory: 'google-de02.jpg'

I've tried both suggestions, but its the same like before. the os.access returns a TRUE but the function cv.LoadImage produce the same error.
Any other solutions?
anyhow thanks fpr answer...

Try giving it the whole path instead of the name only, or maybe using .\google-de02.jpg.

Try using the repr function when the exception happens.